Today I read an interesting generalization of the Rolle's Theorem for Polynomials in $E. 28$ of E. J. Barbeau's book on Polynomials.
It says that if $a, b$ are two consecutive zeroes of polynomial $P(x)$ then the number of roots $P'(x)$ (counting multiplicities) in $(a, b)$ is odd in number. I saw the proof and all.
Now, this made me conjecture a similar thing for any differentiable function, given consecutive roots exist (roots are countable). ALso, I had to assume that the number of roots of $f'(x)$ in $(a, b)$ is finite in number. Also, I don't know if this makes sense, but I also defined multiplicity of roots for $f(x)$ as if $f(a)=f'(a)= \cdots = f^{(n)}(a)=0$ and $f^{(n+1)} \neq 0$, then I say that multiplicity of $a$ is $n$ and I count it as $n$ roots.
My conjecture is that the number of roots of $f'(x)$ (counting orders) stictly between any two consecutive roots of $f$ is odd in number.
Also, I think I have proved it given the assumptions. Because clearly the extremums of $f(x)$ between $a$ and $b$ are odd in number and any point critical point in $(a, b)$ that is not an extremum will be a root with even multiplicity by nth derivative test which I can prove. So, the conjecture holds.
Now, my main questions are, is there a better form of the above conjecture? Also, are any of my assumptions on the countability and infinitude of roots "unecessary"? Are there any analogous results when the roots of $f'(x)$ are infinitude?
Please throw light. Now my question is not very specific or may its asking for too many things, so any related results or links will also be immensely appreciated.
Also I am in high school and don't much analysis analysis (i know uptil spivak calculus).
UPDATE: Since my conjecture (with the assumptions stated seems true) and we are trying to find and analogous extension of it for the case of infinitely many derivatives, I thought we must try to extend the notion of cardinality of a set with odd number of elements to sets with infinitely many elements.
Here is my proposed extension: Call the cardinality of a set $S$ with infinitely many elements, odd iff we cannot cannot partition $S$ into $A$ and $B$ such that there exists a bijection from $A$ to $B$. When the cardinality is finite, then the above definition is clearly true by definition.
So, does the above extension make sense (even when the set is countable)?
Now with the above definition can we say that the number of roots of $f'(x)$ in $(a, b)$ odd? (at least when the roots are countably infinite?)
Define
$$f(x) = \begin{cases} 1-(x-1)^2, & x\le 1 \\ 1+(x-1)^2-(x-1)^3, & x>1 . \end{cases}$$
Then $f$ is continuously differentiable on $\mathbb R,$ is strictly increasing on $(-\infty,5/3)$ and strictly decreasing on $(5/3,\infty).$ Thus $f$ has exactly two zeros, one at $0,$ and one somewhere to the right of $5/3.$ We have $f'(1) = 0 =f'(5/3).$ At neither of these zeros of $f'$ do we have $f'' =0:$ $f''(1)$ doesn't exist, and $f''(5/3) = -2.$ So it would appear the zeros of $f'$ are each of order $1.$ Thus between the consecutive zeros of $f$ we have the number of zeros of $f'$ (counting multiplicity) equal to $2.$
If there is an objection because $f''(1)$ fails to exist, fine. But $f$ is continuously differentiable, hence differentiable, hence satisfies the only hypothesis I see on the smoothness of $f.$ So if this example is disallowed, you need to tell us exactly why, i.e, how smooth $f$ is assumed to be.